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ECE 231 - Resistor Circuit Theory
Contributions by: Ehssan Eizadkhah, HKN Gamma Kappa Member, Co-Founder of ECEBuddy.com Joseph Corleto, HKN Gamma Kappa Member, Co-Founder of ECEBuddy.com __TOC__ Editors Note: This review explains how to calculate resistors that are in series, parallel, or a combination of the two. p-to-T transformation a.k.a ?-to-Y Transformation is also discussed here. Each topic has a solved example. Introduction The resistor is perhaps the simplest component you will ever encounter in your circuit theory class. If you look at any piece of electronic equipment, old or new, chances are there are one or more resistors in there. Current and voltage convention along with two popular symbols for the resistor is shown in Figure 1. What do they exactly do you ask? Simply put, they limit the current and/or voltage in a circuit to a desired level. While doing this, it creates heat, this is the price you have to pay. For most applications this is really not an issue, but for fields such as power electronics and integrated circuit (IC) fabrication, it can pose a real issue. Resistors are manufactured in a variety of values and uses the O (omega) symbol which is the symbol for resistance in the electrical world. Lets say we have a resistor with a value of 100O, we would pronounce this as "one hundred ohms" not "one hundred omegas". They are also non-polar. This means there is no way you can possibly connect them backwards, either direction will do. Series and Parallel Even though there are many values it is sometimes not possible to acquire a certain one from a manufacturer. Or maybe you have only a few kind of resistor values available. How can we get a bunch of resistors and turn them into a higher a lower desired overall value? We can do this by arranging them in a series combination, parallel combination or a mixture of the two. Figure 2 shows three resistors in series. What is the resistance between points A and B if R1 = 1O, R2 = 2O, and R3 = 3O? Well we need to know how to calculate for resistors in series. The formula below tells us how to do this. = + +\ldots + Basically it just says that for resistors in series we just add them up and it gives us the total resistance. Pretty simply right? So for Figure 2 the value between points A and B would be: = + + =1\Omega +2\Omega +3\Omega =6\Omega So this is great if we wanted to get to a value higher than what we have available but how can we go lower? The way would we do this is by simply putting them in parallel. Figure 3 shows how this would look. So what is the value now between points A and B if R1 = 1O, R2 = 2O, and R3 = 3O? The formula below shows us how. =\frac{1}{\frac{1} }+\frac{1} }+\ldots +\frac{1} }} All it says is that we add up the inverse of each resistor, meaning the resistor value divided by 1. After calculating the summation of the inverse of each resistor value, we simply inverse the total. So in our parallel combination above, the resistance between points A and B would be: =\frac{1}{\frac{1} }+\frac{1} }+\frac{1} }} =\frac{1}{\frac{1}{1\Omega }+\frac{1}{2\Omega }+\frac{1}{3\Omega }} =\frac{1}{\frac{11}{6}\Omega }=\frac{6}{11}\Omega \approx 0.545\Omega Notice how the answer turned out to be less than the smallest resistor in the parallel combination, that is, it is less than 1O. By the observation of the parallel formula, this will ALWAYS be the case. So as a sanity check when solving parallel resistors, make sure this is true. It is commonly shown in many texts a simplified formula for two resistors in parallel. If you use the parallel formula shown previously with just two resistors we get the simplified formula below. =\frac } + } Finally we arrive at a situation where series and parallel configurations are together in one circuit. It may look a little intimidating if you are new to circuit theory but after a couple of rounds, it will become second nature. The series-parallel combination is below. What is the resistance between points A and B if R1 = 1O, R2 = 2O, R3 = 3O, and R4 = 4O as shown in Figure 4? Observing the circuit, we can can say that the 1O and the 2O are in series. So these two can be replaced by their series equivalent which would be 3O. Our slightly more simplified circuit is shown in Figure 5. Observing further, we can say that the 3O and the 4O are in parallel. Calculating this we get approximately 1.71O. This answer is also less than the smallest resistor, smaller than the 3O, so it is in a believable range (I did not show you the calculation because you should be calculating along!). So these can be replaced by their parallel equivalent. Our more simplified circuit is shown in Figure 6. Just as a side note the "||" symbol means "in parallel with". It is a commonly used symbol to tell us what things are in parallel with each other. Now there are just two resistors in series. Adding them up we get a total resistance of 3O + 1.71O = 4.71O. The result is shown in Figure 7. Pi-to-T Transformation a.k.a Delta-to-Y Transformation After you have mastered the concept of series and parallel combinations with resistors, you are ready to tackle the p-to-T transformation ("Pi to T"). Basically it takes three resistors that look like a p symbol to something that looks like a T. It is also called a ?-to-Y ("Delta to Wye") transformation because you can change the positioning of the resistors to look like a ? rather than a p and the T to look like a Y; they are equivalent. Figure 8 illustrates how the transformation actually look like. Why on earth would we want to do this? Look at the Figure 9 and see if you can solve for the total resistance between points a and b. Yes, it looks impossible but with the p-to-T conversion, the circuit looks easier. Observe Figures 10a and 10b: Now that is more like it! We can now apply our series and parallel combination formulas to calculate for the total resistance. But how do we even achieve this transformation in the first place? According to Figure that shows the transformation, we have the values of Ra, Rb, and Rc when they are in the p shape so we need to solve for R1, R2, and R3 each as a function of Ra, Rb, and Rc to create the T shape. Let us first derive the equations for the transformation and then we can finish off the problem shown in Figure 9. Now, not going to lie, this is a tedious algebra exercise and it is a little annoying to do. You do not necessarily need to know how to derive the expressions as as long as you memorize the final result of the derivation. So if you are not interested in the details on how the transformation equations came about skip ahead to equations 9, 10, and 11 after the derivation as these are the final equations you need to memorize; they correlate with Figure 8. But if you really want to know how they came about, the derivation is performed below. Go back and observe Figure 8. If we are to do this transformation, the terminal resistance between points a and b have to be the same since they are supposed to be equivalent configurations. This also means that the terminal resistance between points b and c and terminal resistance between points c and also have to be the same. Let us start out by writing the equations for terminal resistance between all points. Referring to Figure 8, we can write: Between points a and b: = \parallel ( + )= + =\frac ( + )} + + }= + Equation 1 Between points b and c: = \parallel ( + )= + =\frac ( + )} + + }= + Equation 2 Between points c and a: = \parallel ( + )= + =\frac ( + )} + + }= + Equation 3 We now have three equations. Let us first solve for R1. We can rewrite Equation 1 and Equation 2 as Equation 4 and Equation 5 respectively. \frac ( + )} + + }- = Equation 4 \frac ( + )} + + }- = Equation 5 Set equations 4 and 5 equal to each other to get equation 6. \frac ( + )} + + }- =\frac ( + )} + + }- Equation 6 Solve Equation 3 for R3 and call it Equation 7. \frac ( + )} + + }- = Equation 7 Substitute Equation 7 into Equation 6 to replace R3. \frac ( + )} + + }- =\frac ( + )} + + }-\frac ( + )} + + }+ Equation 8 Now we have R1, Ra, Rb, and Rc left in the equation. Solve for R1. \frac ( + )} + + }-\frac ( + )} + + }+\frac ( + )} + + }=2 \frac ( + )- ( + )+ ( + )} + + }=2 \frac + - - + + } + + }=2 \frac{2 } + + }=2 \frac } + + }= Equation 9 We have finally figured out what R1 is, we now have to find R2. We can substitute Equation 9 into Equation 4 and figure out the equation for R2. \frac ( + )} + + }-\frac } + + }= \frac + - } + + }= \frac } + + }= Equation 10 We now have an equation for R2. To find R3, we can substitute Equation 9 into Equation 7. \frac ( + )} + + }-\frac } + + }= \frac + - } + + }= \frac } + + }= Equation 11 R3 has finally been solved for and below we summarize the equations that are necessary in order to transform a p configuration into a T configuration. Figure 8 is also shown below for convenience. \frac } + + }= Equation 9 \frac } + + }= Equation 10 \frac } + + }= Equation 11 Now that we have our transformation Equations, we can solve the circuit in Figure 9. As saw in Figures 10a and 10b, the transformation made things pretty easy. By matching our Equation variables to the resistor values in our circuit, we can say that Ra = 37.5O, Rb = 40O, and Rc = 25O. Now we can plug these values into Equations 9, 10, and 11 to solve for R1, R2, and R3 respectively. =\frac } + + }=\frac{40\times 25}{37.5+40+25}\approx 9.76\Omega =\frac } + + }=\frac{37.5\times 25}{37.5+40+25}\approx 9.15\Omega =\frac } + + }=\frac{37.5\times 40}{37.5+40+25}\approx 14.63\Omega Now our circuit looks like Figure 11. Now this becomes an exercise in series-parallel resistors. Let simplify by adding up the 100O and 9.76O and the 125O and 9.15O resistors. Our circuit now looks like Figure 12. Now we can apply the parallel formula for two resistors shown previously. Calculation is below: 109.76||134.15=\frac{109.76\times 134.15}{109.76+134.15}=60.37\Omega Our circuit now looks like Figure 13. Our circuit is now a simple series resistor problem. Adding them up we have approximately 75 Ohms. Conclusion You might be thinking resistors are pretty boring when you are just drilling through problems that only contain them, I completely agree. But it is essential that you work through each problem and understand how to solve them; practice makes perfect. When designing, you are just about always going to need one. They are your friends! As you go on studying and designing circuits your appreciation for them will grow. Category:ECE 231 Category:Circuits Category:Resistors Category:Circuit Theory Category:Pi Category:Transformation Category:Series Category:Parallel